Question

If $\sqrt{\frac{6+2\sqrt{3}}{33-19\sqrt{3}}}=a+b\sqrt{3},$ then $a+b=$

• A. 6
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (B)

8

Given, $\frac{\sqrt{6+2\sqrt{3}}}{33-9\sqrt{3}}=a+b\sqrt{3}$

squaring on both sides, we get,
$(a+b\sqrt{3}{)}^2$ $=\frac{6+2\sqrt{3}}{33-19\sqrt{3}}\times \frac{33+19\sqrt{3}}{33+19\sqrt{3}}$
$=\frac{312+180\sqrt{3}}{6}=52+30\sqrt{3}$
$\Rightarrow a+b\sqrt{3}=\sqrt{52+30\sqrt{3}}$
$=\sqrt{52+2\sqrt{15\times 15\times 3}}$

$=\sqrt{27}+\sqrt{25}$

$=3\sqrt{3}+5$

$\therefore a=5,b=3$

$a+b=8$